How to show that any function of form $z = f(x+at) + g(x-at)$ is a solution of the wave equation
Second partial derivative of $z$ with respect to $t = (a^2)*(\text{second partial derivative of}\,z\,\text {with respect to}\,x)$
I couldn't understand the way of doing it. Can you please help me for this one?
Alternatively you can set $x+at=u$ and $x-at=v$ then $z=f(u) + g(v)$. The wave equation is given by \begin{equation} \frac{\partial ^2z}{\partial t^2} - a^2\frac{\partial ^2z}{\partial x^2} =0, \end{equation} where \begin{equation} \frac{\partial z}{\partial t} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial t} = af'(u) - ag'(v) = a(f'(u)-g'(u)), \end{equation} and \begin{equation} \frac{\partial^2 z}{\partial t^2} = a^2(f''(u) +g''(v)). \end{equation} Similarly for (which you should check) \begin{equation} \frac{\partial^2 z}{\partial x^2} = f''(u) + g''(v). \end{equation} Finally, \begin{equation} \frac{\partial ^2z}{\partial t^2} - a^2\frac{\partial ^2z}{\partial x^2}=a^2(f''(u) +g''(v)) - a^2(f''(u) + g''(v)) =0 \end{equation}
$z = f(x+at)+g(x-at)$, so take the derivatives separately using the chain rule: $$\frac{\partial z}{\partial t} = a f'(x+at) - a g'(x-at)$$ $$\frac{\partial^2 z}{\partial t^2} = a^2 f''(x+at) + a^2 g''(x-at)$$ $$\frac{\partial z}{\partial x} = f'(x+at) + g'(x-at)$$ $$\frac{\partial^2 z}{\partial x^2} = f''(x+at) + g''(x-at)$$ so the result follows.